Connection Between the DTFT and the z-Transform

9.2.2 Connection Between the DTFT and the z-Transform

The connection between the (bilateral) -transform and the DTFT is similar to that between the Laplace transform and the Fourier transform. The -transform of x[n], according to Eq. (5.1) is

Setting = e j Ω in this equation yields

The right-hand side sum defines X( Ω), the DTFT of x[n]. Does this mean that the DTFT can be obtained from the corresponding - transform by setting = e j Ω ? In other words, is it true that X[e j Ω ] = X( Ω)? Yes, it is true in most cases. For example, when x[n] =

a n u[n], its -transform is /( −a), and X[e j Ω ]=e j Ω /(e j Ω − a), which is equal to X(Ω) (assuming |a| < 1). However, for the unit step function u[n], the -transform is /( −1), and X[e j Ω ]=e j Ω /(e j Ω − 1). As seen from Table 9.1 , pair 10, this is not equal to X( Ω) in this case.

We obtained X[e j Ω ] by setting = e j Ω in Eq. (9.42a) . This implies that the sum on the right-hand side of Eq. (9.42a) converges for = e j Ω , which means the unit circle (characterized by = e j Ω ) lies in the region of convergence for X[ ]. Hence, the general rule is that only when the ROC for X[ ] includes the unit circle, setting = e j Ω in X[ ] yields the DTFT X( Ω). This applies for all x[n] that are absolutely summable. If the ROC of X[ ] excludes the unit circle, X[e j Ω ] ≠ X(Ω). This applies to all exponentially growing x[n] and also x[n], which either is constant or oscillates with constant amplitude.

The reason for this peculiar behavior has something to do with the nature of convergence of the -transform and the DTFT. [ † ] This discussion shows that although the DTFT may be considered to be a special case of the -transform, we need to circumscribe such

a view. This cautionary note is supported by the fact that a periodic signal has the DTFT, but its -transform does not exist. [ † ] For the sake of simplicity we assume and therefore X( Ω) to be real. The argument, however, is also valid for complex [or

X( Ω)]. [ † ] This means

[ † ] To explain this point, consider the unit step function u[n] and its transforms. Both the -transform and the DTFT synthesize x[n], using everlasting exponentials of the form n . The value of can be anywhere in the complex -plane for the -transform, but it must be restricted

to the unit circle ( = e j Ω ) in the case of the DTFT. The unit step function is readily synthesized in the -transform by a relatively simple to the unit circle ( = e j Ω ) in the case of the DTFT. The unit step function is readily synthesized in the -transform by a relatively simple

absolutely summable, the region of convergence for the -transform includes the unit circle, and we can synthesize x[n] is absolutely summable, the region of convergence for the -transform includes the unit circle, and we can synthesize x[n] by using along the unit circle in both the transforms. This leads to X[e j Ω ] = X( Ω).